# Thread: convergent sequences

1. x

2. Originally Posted by britnumber
Let X=(x_n) be a sequence of positive real numbers such that = L >1. Show that X is not a bounded sequence and hence is not convergent
Let $\displaystyle \epsilon > 0$ so that $\displaystyle L - \epsilon > 1$.

And therefore there is $\displaystyle N$ so that if $\displaystyle n\geq N$ then $\displaystyle \left| \frac{x_{n+1}}{x_n} - L \right| < \epsilon$.

Therefore, $\displaystyle L - \epsilon< \frac{x_{n+1}}{x_n} \implies (L - \epsilon)x_n < x_{n+1}$.

Therefore,
$\displaystyle (L - \epsilon)x_N < x_{N+1}$
$\displaystyle (L - \epsilon)x_{N+1} < x_{N+2}$
$\displaystyle (L - \epsilon)x_{N+2} < x_{N+3}$
...
Thus,
$\displaystyle (L - \epsilon)^2x_N < x_{N+2}$
$\displaystyle (L - \epsilon)^3x_N < x_{N+3}$
...
In general,
$\displaystyle (L - \epsilon)^kx_N < x_{N+k}$

But $\displaystyle (L - \epsilon)^k$ can be made arbitary large since $\displaystyle L - \epsilon > 1$.
Therefore, $\displaystyle \{ x_n\}$ is unbounded.